Question

Тригонометрия 10 класс ,помогите решить уравнение

$2sin^{2} x+\sqrt{3} cosx=\sqrt{6}cos(x-\frac{\pi}{4})$

Answer 1

$2sin^2x+\sqrt{3}*cosx=\sqrt{6}*cos\bigg(x-\dfrac{\pi}{4}\bigg)\\\\\boldsymbol{*}\:\:\:cos\bigg(x-\dfrac{\pi}{4}\bigg)=cosx*cos\dfrac{\pi}{4}+sinx*sin\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}*\Big(sinx+cosx\Big)\\\\Formyla\::\\\\\boxed{\:\:cos\Big(\alpha-\beta\Big)=cos\alpha*cos\beta+sin\alpha*sin\beta\:\:}$

$2sin^2x+\sqrt{3}*cosx=\sqrt{6}*\dfrac{\sqrt{2}}{2}*\Big(sinx+cosx\Big)\\\\2sin^2x+\sqrt{3}*cosx=\sqrt{3}*\Big(sinx+cosx\Big)\\\\2sin^2x+\sqrt{3}*cosx=\sqrt{3}*sinx+\sqrt{3}*cosx\\\\2sin^2x=\sqrt{3}*sinx\\\\sinx*\Big(\:2sinx-\sqrt{3}\:\Big)=0\\\\sinx*\Big(\:sinx-\dfrac{\sqrt{3}}{2}\:\Big)=0\\\\1)\:\:\:sinx=0\\\\\boxed{\:\:x\:_k=\pi\:k\:\:,\:\:\:k\:\in\:\mathbb{Z}\:\:}$

$2)\:\:\:sinx-\dfrac{\sqrt{3}}{2}=0\\\\sinx=\dfrac{\sqrt{3}}{2}\\\\\boxed{\:\:x\:_m=\dfrac{\pi}{3}+2\pi\:m\:\:,\:\:\:m\:\in\:\mathbb{Z}\:\:}\\\\\\\boxed{\:\:x\:_t=\dfrac{2\pi}{3}+2\pi\:t\:\:,\:\:\:t\:\in\:\mathbb{Z}\:\:}$